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KVL and KCL
KVL stands for Kirchhoff voltage law. KVL states that the total voltage drops around the loop in any closed electric circuit is equal to the sum of total voltage drop in the same closed loop.
Sign Convention
Science and technology incorporate some ideas and techniques of their own to understand a system skilfully and easily. These techniques are called conventions. For example: Sign conventions of mirrors are used to understand the phenomenon of reflection and refraction in an easier way.
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- Part b.) (cutt off) repeat using nodal analysis. (Find phaser voltage at above values)Three single-phase two-winding transformers, each rated 3 kVA, 220/110volts, 60 Hz, with a 0.10 per-unit leakage reactance, are connected as athree-phase extended D autotransformer bank, as shown in Figure 3.36 (c).The low-voltage D winding has a 110-volt rating. (a) Draw the positive sequencephasor diagram and show that the high-voltage winding has a479.5-volt rating. (b) A three-phase load connected to the low-voltageterminals absorbs 6 kW at 110 volts and at 0.8 power factor lagging. Drawthe per-unit impedance diagram and calculate the voltage and current atthe high-voltage terminals. Assume positive-sequence operation.HOME WORK: Per-unit circuit: three-zone single-phase network determine the per-unit impedances and the per-unit source voltage. Then cal- culate the load current both in per-unit and in amperes. Transformer winding resistances and shunt admittance branches are neglected. Zone 1 Vs = 220/0° volts i Three zones of a single-phase circuit are identified in Figure 3.10(a). The zones are connected by transformers T₁ and T₂, whose ratings are also shown. Using base values of 30 kVA and 240 volts in zone 1, draw the per-unit circuit and T₁ 30 kVA 240/480 volts Xeq = 0.10 p.u. V sp.u. = 0.9167/0° p.u. Zoase Zone 2 Xune 20 (a) Single-phase circuit I'spu i Xtio.u Zone 1 Vbase1 = 240 volts (240)² 30,000 Spase = 30 kVA = 1.92 T₂ Zload = 0.9 +0.2 20 KVA 460/115 volts Xea = 0.10 p.u. jXunepu j0.10 p.u. j0.2604 p.u. Zone 2 Vbase2 = 480 volts Zbase2 = Zone 3 (480)² 30,000 ¡XT2pu hoadp.u. = 7.68 (b) Per-unit circuit /0.1378 p.u. Zone 3 Vbase3= 120 volts Zbase3 = base3 = Zoadp.u. = 1.875+ 0.4167 p.u.…
- Consider three ideal single-phase transformers (with a voltage gain of ) put together as three-phase bank as shown in Figure 3.35. Assuming positive-sequence voltages for Va,Vb, and Vc find Va,Vb, and VC. in terms of Va,Vb, and Vc, respectively. (a) Would such relationships hold for the line voltages as well? (b) Looking into the current relationships, express IaIb and Ic in terms of IaIb and Ic respectively. (C) Let S and S be the per-phase complex power output and input. respectively. Find S in terms of S.Consider the oneline diagram shown in Figure 3.40. The three-phase transformer bank is made up of three identical single-phase transformers, each specified by X1=0.24 (on the low-voltage side), negligible resistance and magnetizing current, and turns ratio =N2/N1=10. The transformer bank is delivering 100 MW at 0.8 p.f. lagging to a substation bus whose voltage is 230 kV. (a) Determine the primary current magnitude, primary voltage (line-to-line) magnitude, and the three-phase complex power supplied by the generator. Choose the line-to-neutral voltage at the bus, Va as the reference Account for the phase shift, and assume positive-sequence operation. (b) Find the phase shift between the primary and secondary voltages.Three single-phase two-winding transformers, each rated 25MVA,54.2/5.42kV, are connected to form a three-phase Y- bank with a balanced Y-connected resistive load of 0.6 per phase on the low-voltage side. By choosing a base of 75 MVA (three phase) and 94 kV (line-to-line) for the high-voltage side of the transformer bank, specify the base quantities for the low-voltage side. Determine the per-unit resistance of the load on the base for the low-voltage side. Then determine the load resistance RL in ohms referred to the high-voltage side and the per-unit value of this load resistance on the chosen base.
- 3.6 For a conceptual single-phase phase-shifting transformer, the primary voltage leads the secondary voltage by 30°. A load connected to the sec- ondary winding absorbs 110 kVA at an 0.8 power factor leading and at a voltage E, = 277/0° volts. Determine (a) the primary voltage, (b) primary and secondary currents, (c) load impedance referred to the primary wind- ing, and (d) complex power supplied to the primary winding.b) A fault occurs at bus 4 of the network shown in Figure Q3. Pre-fault nodal voltages throughout the network are of 1 p.u. and the impedance of the electric arc is neglected. Sequence impedance parameters of the generator, transmission lines, and transformer are given in Figure Q3, where X and Y are the last two digits of your student number. V₁ = 120° p.u. V₂ = 120° p.u. jX(1) j0.1Y p.u. jX2)= j0.1Y p.u. jXko) j0.1X p.u. - 0 jX(1) = j0.2 p.u. 1JX(2) = 0.2 p.u. 2 jX1(0) = j0.25 p.u. jX2(1) j0.2 p.u. V₁=1/0° p.u. jX(2(2) = j0.2Y p.u. jX2(0) = j0.3X p.u. = V₂ = 120° p.u. jXT(1) j0.1X p.u. jXT(2) j0.1X p.u. JX3(1) j0.1Y p.u. JX3(2)=j0.1Y p.u. jXT(0) j0.1X p.u. JX3(0)=j0.15 p.u. 0- = 3 = Figure Q3. Circuit for problem 3b). For example, if your student number is c1700123, then: jXa(n) = j0.13 p. u., jXa(z) = j0.13 p. u., and jXa(o) = j0.12 p. u. 4 (i) Assuming a balanced excitation, draw the positive, negative and zero sequence Thévenin equivalent circuits as seen from bus 4. (ii)…b) A fault occurs at bus 3 of the network shown in Figure Q4. Pre-fault nodal voltages throughout the network are of 1 p.u. and the impedance of the electric arc is neglected. Sequence impedance parameters of the generator, transmission lines, transformer and load are given in Figure Q4. V₁ = 120° p.u. V₂ = 120° p.u. V₂ = 1/0° p.u. V₂= 120° p.u. jXj0.1 p.u. JX2) 0.1 p.u. jX0j0.15 p.u. jXn-j0.2 p.u. 1 JX(2)-j0.2 p.u. 2 jX)=j0.25 p.u. JX20-10.15 p.u. jXa(z)-j0.2 p.u. 4 jX2(0)=j0.2 p.u. jXT(1) j0.1 p.u. jXT(2)=j0.15 p.u. jXT(0)=j0.1 p.u. Figure Q4. Circuit for problem 4b). = jXj0.1 p.u. j0.1 p.u. - JX(2) JXL(0) 10.1 p.u. = (i) Assuming a balanced excitation, draw the positive, negative and zero sequence Thévenin equivalent circuits as seen from bus 3. (ii) Determine the positive sequence fault current for the case when a three- phase-to-ground fault occurs at bus 3 of the network. (iii) Determine the short-circuit fault current for the case when a one-phase- to-ground fault occurs at bus…
- A system of three currents equal in magnitude and having zero phase displacement, these are called. Select one: a. negative phase sequence components b. zero phase sequence components c. None of above d. positive phase sequence components.A system of three currents equal in magnitude and having zero phase displacement, these are called. Select one: a. None of above b. negative phase sequence components c. positive phase sequence components. d. zero phase sequence componentsQ2) In the network in the figure below Y-Y connected transformers, each with grounded neutrals, are at the ends of each transmission line that is not terminating at bus 3. The transformers connecting the lines to bus 3 are Y-A, with the neutral of the Y solidly grounded and the A sides connected to bus 3. All the line reactances shown in the figure between busses include the reactances of the transformers. Zero sequence values for these lines including transformers are 2.0 times those shown in the figure. Both generators are Y-Connected. Zero-sequence reactances of the generators connected to bus 1 and bus 3 are 0.04 and 0.08 per unit, respectively. The neutral of the generator at bus 1 is connected to ground through a reactor of 0.02 per unit; the generator at bus 3 has a solidly ground neutral. Find the bus impedance matrices (¹), (²), z for the given network and 'bus' 'bus' bus then compute the Subtransient current in per unit for a single line-to-ground fault on bus 2 and the fault…